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This is a physics problem about finding the electric field produced by two point charges. The problem gives us two charges, each with charge \(+Q\), placed at the ends of a rhombus-shaped figure. The two relevant points for calculation are point \(P\) and point \(P'\). We are tasked with finding the electric field at these points due to the two charges.

### Step-by-step solution approach:

#### 1. **Electric Field at Point \(P\):**
   - The electric field at any point due to a charge is given by:
     \[
     \vec{E} = \frac{kQ}{r^2} \hat{r}
     \]
     where \(k\) is Coulomb's constant, \(Q\) is the charge, \(r\) is the distance from the charge to the point, and \(\hat{r}\) is the unit vector pointing from the charge to the point.

   - At point \(P\), the electric field due to the charge at the opposite side can be calculated by symmetry. Since the charges are equal and positioned symmetrically, their horizontal components will cancel out, and only the vertical components of the electric fields will add up.

   - Each charge produces an electric field of magnitude:
     \[
     E = \frac{kQ}{d^2}
     \]
     The vertical component of this field is:
     \[
     E_y = E \sin(45^\circ) = \frac{kQ}{d^2} \frac{1}{\sqrt{2}}
     \]
     Since there are two charges contributing, the total electric field at point \(P\) is:
     \[
     E_{total} = 2E_y = 2 \times \frac{kQ}{d^2} \frac{1}{\sqrt{2}} = \frac{\sqrt{2}kQ}{d^2}
     \]

#### 2. **Electric Field at Point \(P'\):**
   - For point \(P'\), the charges are located at different distances. The distance from the charge on the left to point \(P'\) is \(d\), and the distance from the charge on the right to point \(P'\) is \(2d\).

   - The electric field at point \(P'\) due to the charge at distance \(d\) is:
     \[
     E_1 = \frac{kQ}{d^2}
     \]
     The electric field at point \(P'\) due to the charge at distance \(2d\) is:
     \[
     E_2 = \frac{kQ}{(2d)^2} = \frac{kQ}{4d^2}
     \]
     Both electric fields need to be broken into components. The vertical components from each charge will add up similarly to point \(P\).

     Using the geometry and summing components, the final electric field can be determined by vector addition of these fields.

Would you like me to calculate these electric field values numerically or provide further details on the geometry and vector breakdown? 

### Relative questions:
1. How does the distance between two charges affect the electric field strength at a point?
2. Why do the horizontal components of the electric field at point \(P\) cancel out?
3. How would the electric field change if one of the charges was negative?
4. What role does symmetry play in simplifying the calculation of the electric field in this scenario?
5. Could this approach be used for calculating the potential at points \(P\) and \(P'\)?

### Tip:
Always use symmetry whenever possible in electric field problems, as it often simplifies the component analysis significantly.

Find the electric field at points P and P' due to two equal charges placed at the corners of a rhombus, as shown in the image.

Calculate the electric field at points P and P' due to two point charges located at the corners of a rhombus. This problem covers key electromagnetism concepts, such as vector addition and the use of Coulomb's law.

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